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2022 AP Calculus AB FRQ Question 5: A Differential Equation with a Square-Root Factor

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2022.

Question 5

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A Differential Equation with a Square-Root Factor

Differential Equations

Hard
Consider the differential equation dydx=12sin⁡ ⁣(π2x)y+7\dfrac{dy}{dx}=\dfrac{1}{2}\sin\!\left(\dfrac{\pi}{2}x\right)\sqrt{y+7}. Let y=f(x)y=f(x) be the particular solution to the differential equation with the initial condition f(1)=2f(1)=2. The function ff is defined for all real numbers.
Part AEasy1 point
A portion of the slope field for the differential equation is given. Sketch the solution curve through the point (1,2)(1,2).

Answer

The curve rises through (1,2)(1,2), following the slope field.
Full Solution & Work

Follow the slope field from (1, 2)

At (1,2)(1,2): dydx=12sin⁡(π/2)9=12(1)(3)=1.5\frac{dy}{dx}=\frac12\sin(\pi/2)\sqrt9=\frac12(1)(3)=1.5, a fairly steep positive slope. Following the field, the curve continues rising to the right and falls off to the left of x=1x=1, consistent with the shown slope field.

AP Scoring — 1 Point

**P1**: The solution curve must pass through (1,2)(1,2) and have no obvious conflicts with the given slope lines.
Part BMedium2 points
Write an equation for the line tangent to the solution curve in part (a) at the point (1,2)(1,2). Use the equation to approximate f(0.8)f(0.8).

Answer

y=2+1.5(x−1)y=2+1.5(x-1); f(0.8)≈1.7f(0.8)\approx1.7
Full Solution & Work

Find the slope at (1,2)

dydx∣(1,2)=12sin⁡ ⁣(π2)2+7=12(1)(3)=1.5\left.\frac{dy}{dx}\right|_{(1,2)} = \frac12\sin\!\left(\frac{\pi}{2}\right)\sqrt{2+7} = \frac12(1)(3)=1.5

Write the tangent line and approximate

y=2+1.5(x−1)y=2+1.5(x-1)

f(0.8)≈2+1.5(0.8−1)=2−0.3=1.7f(0.8) \approx 2+1.5(0.8-1)=2-0.3=1.7

AP Scoring — 2 Points

**P1**: Correct slope, 1.5, at (1,2)(1,2).
**P2**: Correct approximation, f(0.8)≈1.7f(0.8)\approx1.7.
Part CMedium1 point
It is known that f′′(x)>0f''(x)>0 for −1≤x≤1-1\leq x\leq1. Is the approximation found in part (b) an overestimate or an underestimate for f(0.8)f(0.8)? Give a reason for your answer.

Answer

Underestimate.
Full Solution & Work

Relate concavity to tangent-line approximation

Since f′′(x)>0f''(x)>0 on [−1,1][-1,1], the graph of ff is **concave up** there, so the tangent line lies **below** the curve on this interval.

Conclude

Since 0.8∈[−1,1]0.8\in[-1,1], the tangent-line approximation f(0.8)≈1.7f(0.8)\approx1.7 is an **underestimate** of the actual value.

AP Scoring — 1 Point

**P1**: "Underestimate," with reasoning tied to f′′(x)>0f''(x)>0 (concave up).
Part DHard3 points
Use separation of variables to find y=f(x)y=f(x), the particular solution to the differential equation dydx=12sin⁡ ⁣(π2x)y+7\dfrac{dy}{dx}=\dfrac12\sin\!\left(\dfrac{\pi}{2}x\right)\sqrt{y+7} with the initial condition f(1)=2f(1)=2.

Answer

y=[3−12πcos⁡ ⁣(π2x)]2−7y = \left[3-\dfrac{1}{2\pi}\cos\!\left(\dfrac{\pi}{2}x\right)\right]^2-7
Full Solution & Work

Separate variables

dyy+7=12sin⁡ ⁣(π2x)dx\frac{dy}{\sqrt{y+7}} = \frac12\sin\!\left(\frac{\pi}{2}x\right)dx

Integrate both sides

2y+7=−1πcos⁡ ⁣(π2x)+C2\sqrt{y+7} = -\frac{1}{\pi}\cos\!\left(\frac{\pi}{2}x\right)+C

Apply the initial condition f(1) = 2

At x=1,y=2x=1,y=2: 29=6=−1πcos⁡(π/2)+C=0+C  ⟹  C=62\sqrt9=6=-\frac1\pi\cos(\pi/2)+C=0+C \implies C=6.
2y+7=6−1πcos⁡ ⁣(π2x)2\sqrt{y+7} = 6-\frac{1}{\pi}\cos\!\left(\frac\pi2 x\right)

Solve for y

y+7=3−12πcos⁡ ⁣(π2x)\sqrt{y+7} = 3-\frac{1}{2\pi}\cos\!\left(\frac\pi2 x\right)

y=[3−12πcos⁡ ⁣(π2x)]2−7y = \left[3-\frac{1}{2\pi}\cos\!\left(\frac\pi2 x\right)\right]^2-7

AP Scoring — 3 Points

**P1**: Correct separation of variables.
**P2**: Correct antiderivatives on both sides, with constant of integration correctly evaluated using f(1)=2f(1)=2.
**P3**: Correctly solves for yy to reach the final closed form.

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